Without multiplying matrices, find bases for the row and column spaces of A:
$$A= \begin{bmatrix} 1 & 2 \\ 4 & 5 \\ 2 & 7 \\ \end{bmatrix} \begin{bmatrix} 3 & 0 & 3 \\ 1 & 1 & 2 \\ \end{bmatrix} $$

How do you know from these shapes tha A cannot be invertible?

I don't know how to figure this out without multiplying?
I think the bases can be constructed by the linear combinations of the matrices where the numbers of the second matrix are the coefficients, something like this would give us the first column in $A$ after multiplication: $$3\begin{bmatrix} 1 \\ 4 \\ 2 \\ \end{bmatrix} +1\begin{bmatrix} 2 \\ 5 \\ 7 \\ \end{bmatrix}$$


Since the first matrix has rank $2$, the rank of the product $A$ will be at most $2$. Since $A$ is a $3 \times 3$ matrix, it will not be invertible.

As to the column space, you are right, it will consist of all the linear combinations of the two columns of the left matrix.

The argument for the row space is similar to the argument you are using for the columns. Note that the first row of $A$ will be $$1 \cdot \begin{bmatrix} 3 & 0 & 3 \\ \end{bmatrix} + 2 \cdot \begin{bmatrix} 1 & 1 & 2 \\ \end{bmatrix}$$

Note that in determining the column and row spaces you are also proving the first statement about the rank.

  • $\begingroup$ the rank of the first matrix is 2 and the rank of the second matrix is also 2, that means the rank of A is also 2 and so is the dimension of the row and column space of A, that means I need two vectors for the bases of the column space and the bases of the row space, but I don't see how I can construct a bases without multipying and finding the pivot columns? $\endgroup$ – idknuttin Mar 19 '16 at 17:02
  • $\begingroup$ Look at the column space. You have given one of its elements, the first column, as a linear combination of the two columns of the left matrix. Look at the second column of $A$ - it will be a different linear combination of the same two two columns of the left matrix. This tells you that these two columns are a base for the column space. $\endgroup$ – Andreas Caranti Mar 19 '16 at 17:23
  • $\begingroup$ so would the bases of the column space of A be $(5\qquad17\qquad13)$ and $(2\qquad5\qquad7)$? $\endgroup$ – idknuttin Mar 19 '16 at 18:47

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